📚 Year 10 SQA Advanced Mathematics: Case Study Practical Exercises | Year 10 SQA 进阶数学：案例分析实战演练
In the SQA Advanced Mathematics syllabus, applying algebraic, trigonometric and geometric techniques to real‑world scenarios is a core skill. This article presents a series of practical case studies that mirror the type of reasoning and problem‑solving expected at Year 10. Each example is broken down into clear logical steps, with English and Chinese explanations provided side by side to reinforce understanding.
在 SQA 进阶数学的课程中，将代数、三角和几何技巧应用到现实场景是一项核心能力。本文通过一系列实践案例演练，模拟了 Year 10 阶段常见的推理与解题思维。每个实例都按清晰的逻辑步骤分解，并提供中英文对照说明，帮助大家巩固理解。
1. Quadratic Modelling – Fountain Trajectory | 案例一：二次函数建模——喷泉轨迹
A decorative fountain shoots water from ground level. The jet reaches a maximum height of 3 m when it is 4 m horizontally from the nozzle and then lands 8 m away. Model the parabolic path with a quadratic function and find the height 2 m from the start.
一座装饰喷泉从地面向上喷水。水柱在水平距离喷嘴 4 m 处达到最高 3 m，随后在 8 m 处落地。请用二次函数建立抛物线轨迹，并求出离起点 2 m 处的高度。
Step 1: Use the vertex form y = a(x − 4)² + 3, since the vertex is at (4, 3).
步骤一：使用顶点式 y = a(x − 4)² + 3，因为顶点坐标为 (4, 3)。
Step 2: Substitute the launch point (0, 0) to find a: 0 = a(0 − 4)² + 3 → 16a = −3 → a = −3/16.
步骤二：代入起点 (0, 0) 求 a：0 = a(0 − 4)² + 3 → 16a = −3 → a = −3/16。
y = −3/16 (x − 4)² + 3
Step 3: Evaluate at x = 2: y = −3/16 (2 − 4)² + 3 = −3/16 × 4 + 3 = −12/16 + 3 = 2.25 m.
步骤三：当 x = 2 时，y = −3/16 (2 − 4)² + 3 = −3/16 × 4 + 3 = −12/16 + 3 = 2.25 m。
2. Trigonometric Heights – Hillside Survey | 案例二：三角测高——山坡勘测
A surveyor measures the angle of elevation from a point 100 m away from the base of a hill and obtains 30°. She then walks 50 m closer to the base and measures an elevation of 45°. Determine the height of the hill above the measurement point.
测量员在离山脚 100 m 处测得仰角为 30°。她又朝山脚走近 50 m，测得仰角为 45°。试求出测量点到山顶的高度。
Let h be the height and d e second observation point to the point directly below the summit. From the 45° angle: tan 45° = h/d → h = d.
设 h 为高度，d 为第二个观察点到山顶垂足的水平距离。由 45° 角可得：tan 45° = h/d → h = d。
From the first observation: tan 30° = h/(d + 50). Substitute h = d: 1/√3 = h/(h + 50).
从第一个观测点可得：tan 30° = h/(d + 50)。将 h = d 代入：1/√3 = h/(h + 50)。
Cross‑multiply: h + 50 = h√3 → h√3 − h = 50 → h(√3 − 1) = 50 → h = 50/(√3 − 1) ≈ 68.3 m.
交叉相乘：h + 50 = h√3 → h√3 − h = 50 → h(√3 − 1) = 50 → h = 50/(√3 − 1) ≈ 68.3 m。
3. Simultaneous Equations – Alloy Blending | 案例三：联立方程组——合金配比
A foundry has two batches of copper alloy: one containing 30 % copper and the other 60 % copper. How many kilograms of each must be mixed to produce 100 kg of an alloy that is 45 % copper?
某铸造厂有两种铜合金：一种含铜 30%，另一种含铜 60%。需要混合多少千克每种合金才能制得 100 kg 含铜 45% 的合金？
Let x be the mass of the 30 % alloy and y be the mass of the 60 % alloy. The equations are:
设 x 为 30% 合金的质量，y 为 60% 合金的质量。方程组为：
x + y = 100
0.30x + 0.60y = 45
Multiply the second equation by 100: 30x + 60y = 4500 → divide by 30: x + 2y = 150.
将第二个方程乘以 100：30x + 60y = 4500 → 除以 30：x + 2y = 150。
Subtract x + y = 100 from x + 2y = 150: y = 50, then x = 50. Therefore, 50 kg of each is required.
用 x + 2y = 150 减去 x + y = 100，得 y = 50，继而 x = 50。因此需要各 50 kg。
4. Exponential Growth – Bacteria Colony | 案例四：指数增长——细菌菌落
urs. Initially there are 500 bacteria, and after 3 hours the count is 4000. Calculate the growth constant k and predict the number after 5 hours.
一菌落遵循模型 N = N₀eᵏᵗ，N 为 t 小时后的数量。初始有 500 个细菌，3 小时后增至 4000。求增长常数 k，并预测 5 小时后的数量。
Using the given data: 4000 = 500 e³ᵏ → e³ᵏ = 8 → 3k = ln 8 → k = (ln 8)/3 ≈ 0.6931.
代入已知数据：4000 = 500 e³ᵏ → e³ᵏ = 8 → 3k = ln 8 → k = (ln 8)/3 ≈ 0.6931。
After 5 hours: N = 500 e⁵ᵏ = 500 e5 ln 8 / 3 = 500 × 85/3 = 500 × (2³)5/3 = 500 × 2⁵ = 500 × 32 = 16000.
5 小时后：N = 500 e⁵ᵏ = 500 e5 ln 8 / 3 = 500 × 85/3 = 500 × (2³)5/3 = 500 × 2⁵ = 500 × 32 = 16000。
5. Vectors – River Crossing | 案例五：向量——横渡河流
A river flows due east at 3 m s⁻¹. A boat, capable of 5 m s⁻¹ in still water, aims to travel directly north across the river. Find the heading the boat must steer and its resultant speed.
一条河以 3 m s⁻¹ 的速度向东流动。一艘船在静水中航速为 5 m s⁻¹，想直接向北横渡。求船应采取的航向角及实际速度。
tilted westwards. The eastward component of the boat’s velocity must cancel the current: 5 sin θ = 3 → sin θ = 0.6 → θ ≈ 36.87° west of north.
设 θ 为船头与正北方向的夹角，偏向西。船速的东西分量需抵消水流：5 sin θ = 3 → sin θ = 0.6 → θ ≈ 36.87°，北偏西。
Resultant northward speed: v = 5 cos θ = 5 × 0.8 = 4 m s⁻¹ (or by Pythagoras: √5² − 3² = 4).
实际向北速度：v = 5 cos θ = 5 × 0.8 = 4 m s⁻¹（或用勾股定理：√5² − 3² = 4）。
6. Arithmetic & Geometric Sequences – Saving Plan | 案例六：数列——储蓄计划
At the beginning of each month, £100 is deposited into an account paying 6 % per annum compounded monthly. How much will be in the account at the end of one year, just after the 12th deposit?
每月初存入 £100，年利率 6%，每月复利。问一年末（即第 12 次存款刚存入后）账户里有多少钱？
The monthly interest rate is 0.5 % . The value of the 1st deposit after 12 months is 100 × 1.005¹², the 2nd is 100 × 1.005¹¹, …, the 12th is 100 × 1.005. This forms a geometric progression.
月利率为 0.5%（增长因子 1.005）。第一笔存款 12 个月后价值为 100 × 1.005¹²，第二笔为 100 × 1.005¹¹，……，第十二笔为 100 × 1.005。这构成一个等比数列。
puting: 1.005¹² ≈ 1.06168, so S ≈ 100 × 1.005 × (0.06168 / 0.005) ≈ 100 × 1.005 × 12.336 ≈ £1239.77.
总和 S = 100 × 1.005 (1.005¹² − 1) / (1.005 − 1) = 100 × 1.005 × (1.005¹² − 1) / 0.005。计算：1.005¹² ≈ 1.06168，所以 S ≈ 100 × 1.005 × (0.06168 / 0.005) ≈ 100 × 1.005 × 12.336 ≈ £1239.77。
7. Quadratic Optimisation – Maximum Fenced Area | 案例七：二次函数最优化——最大围栏面积
A farmer has 100 m of fencing to enclose a rectangular plot against a long straight wall. What dimensions give the maximum area, and what is that area?
一位农场主有 100 m 长的围栏，想靠着一条长直墙壁围成一个矩形地块。问什么尺寸能使面积最大？最大面积是多少？
wall) and length L (parallel to the wall). The fencing covers L + 2x = 100 → L = 100 − 2x.
设地块宽为 x（垂直于墙），长为 L（平行于墙）。围栏总长为 L + 2x = 100 → L = 100 − 2x。
= −2x² + 100x. This quadratic opens downwards; maximum at x = −b/(2a) = −100/(2·(−2)) = 25 m.
面积 A = x · L = x(100 − 2x) = −2x² + 100x。该二次函数开口向下，最大值在 x = −b/(2a) = −100/(2·(−2)) = 25 m。
Then L = 100 − 2(25) = 50 m. Maximum area = 25 × 50 = 1250 m².
于是 L = 100 − 2(25) = 50 m。最大面积 = 25 × 50 = 1250 m²。
8. 3D Pythagoras – Space Diagonal | 案例八：三维勾股定理——空间对角线
A storage box measures 10 cm long, 6 cm wide and 8 cm high. Calculate the length of the longest rod that can fit straight inside the box.
一个储物箱长 10 cm、宽 6 cm、高 8 cm。求能笔直放入箱内的最长杆长度。
The space diagonal d of a cuboid is given by d = √l² + w² + h². Substituting: d = √10² + 6² + 8² = √100 + 36 + 64 = √200.
长方体的空间对角线 d 由 d = √l² + w² + h² 给出。代入得：d = √10² + 6² + 8² = √100 + 36 + 64 = √200。
Simplify: √200 = √100 × 2 = 10√2 ≈ 14.14 cm.
化简：√200 = √100 × 2 = 10√2 ≈ 14.14 cm。
9. Trigonometric Equations – Solving for Angles | 案例九：三角方程——求解角度
0° ≤ θ < 360°. This equation appears when analysing alternating stresses in materials.
在 0° ≤ θ < 360° 范围内解方程 2 sin² θ − sin θ − 1 = 0。该方程常用于材料交变应力的分析。
sin θ: 2X² − X − 1 = 0 where X = sin θ. Factorise: (2X + 1)(X − 1) = 0 → X = −½ or X = 1.
视作关于 sin θ 的二次方程：2X² − X − 1 = 0，其中 X = sin θ。分解因式：(2X + 1)(X − 1) = 0 → X = −½ 或 X = 1。
= 90°. For sin θ = −½: reference angle 30°, so θ = 180° + 30° = 210° or 360° − 30° = 330°.
当 sin θ = 1 时，θ = 90°。当 sin θ = −½ 时，参考角为 30°，故 θ = 180° + 30° = 210° 或 360° − 30° = 330°。
Solution set: θ = 90°, 210°, 330°.
解集为：θ = 90°, 210°, 330°。
10. Composite Functions – Modelling Costs | 案例十：复合函数——成本建模
A factory applies a 20 % discount on the raw price, then adds a £15 delivery charge. Write a composite function for the final cost and evaluate it for a raw price of £200.
一家工厂对原价给予 20% 的折扣，然后加上 £15 运费。写出最终费用的复合函数，并计算原价为 £200 时的费用。
= 0.8x represent the discount and g(x) = x + 15 the delivery charge. The final cost is (g ∘ f)(x) = g(f(x)) = 0.8x + 15.
设 f(x) = 0.8x 代表折扣后的价格，g(x) = x + 15 代表加上运费。最终费用为 (g ∘ f)(x) = g(f(x)) = 0.8x + 15。
For x = 200: final cost = 0.8 × 200 + 15 = 160 + 15 = £175.
当 x = 200 时，最终费用 = 0.8 × 200 + 15 = 160 + 15 = £175。
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